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(D) After a long time,the capacitor is fully charged and acts as an open circuit. Therefore,no current flows through the branch containing the capacit

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(D) After a long time,the capacitor is fully charged and acts as an open circuit. Therefore,no current flows through the branch containing the capacitor.The equivalent circuit consists of the $6 \, V$ battery in series with the $1 \, k\Omega$ and $2 \, k\Omega$ resistors.The total resistance of the circuit is $R_{eq} = 1 \, k\Omega + 2 \, k\Omega = 3 \, k\Omega$.The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{6 \, V}{3 	imes 10^3 \, \Omega} = 2 	imes 10^{-3} \, A = 2 \, mA$.The voltage across the $2 \, k\Omega$ resistor is $V_{AB} = I 	imes R = (2 	imes 10^{-3} \, A) 	imes (2 	imes 10^3 \, \Omega) = 4 \, V$.Since the capacitor is in parallel with the $2 \, k\Omega$ resistor,the voltage across the capacitor is equal to the voltage across the $2 \, k\Omega$ resistor,which is $4 \, V$.

The circuit shown has been connected for a long time. The voltage across the capacitor is ............. $V$.

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